Question

Three numbers are chosen at random without replacement from $1,2,3,...,10$. The probability that the minimum of the chosen numbers is $4$ or their maximum is $8$, is

• A. $\frac{11}{40}$
• B. $\frac{3}{10}$
• C. $\frac{1}{40}$
• D. none of these

Answer 1

The correct option is A $\frac{11}{40}$

$P$ (min $4$ or max $8$) = P(min $4$) + P(max $8$) - P(min $4$ and max $8$)
P(min $4$) $=6C_2/10C_3$
P(max 8) $=7C_2/10C_3$
P(min $4$ and max $8$)$=3C_1/10C_3$
$P=15/120+21/120-3/120=11/40$